Question

A solenoid of \(1.5\) meter length and \(4 \mathrm{~cm}\) diameter possesses 10 turn per \(\mathrm{cm}\). A current of \(5 \mathrm{Amp}\) is flowing through it. The magnetic induction at axis inside the solenoid is (a) \(2 \pi \times 10^{-3} \mathrm{~T}\) (b) \(2 \pi \times 10^{-5} \mathrm{~T}\) (c) \(2 \pi \times 10^{-2} \mathrm{G}\) (d) \(2 \pi \times 10^{-5} \mathrm{G}\)

Short answer

The magnetic induction at the axis inside the solenoid is \(B = 2\pi \times 10^{-3}\ \mathrm{T}\).

Step-by-step solution

Identify the given parameters

We are given the length of the solenoid \(l\), diameter \(d\), number of turns per centimeter \(n\), and the current flowing through the solenoid \(I\). These values are: Length of solenoid (\(l\)) = \(1.5\ \mathrm{m}\) Diameter of solenoid (\(d\)) = \(4\ \mathrm{cm}\) Turns per centimeter (\(n\)) = \(10\ \mathrm{turns/cm}\) Current through solenoid (\(I\)) = \(5\ \mathrm{A}\)

Calculate the total number of turns in the solenoid

To find the total number of turns in the solenoid, we must multiply the number of turns per centimeter by the length of the solenoid in centimeters: Total number of turns (\(N\)) = \(n \times l\) We have \(l = 1.5\ \mathrm{m} = 150\ \mathrm{cm}\) (converting meters to centimeters). So: \(N = 10\ \mathrm{turns/cm} \times 150\ \mathrm{cm} = 1500\ \mathrm{turns}\)

Apply Ampere's Law to find magnetic field (induction) B

Ampere's law states that the magnetic field (induction) inside a solenoid can be calculated as: \(B = \mu_0 n I \) Here, \(\mu_0\) is the permeability of free space, and its value is \( 4\pi \times 10^{-7}\ \mathrm{Tm/A} \) \(n\) is the density of turns (turns per meter) and requires conversion from turns per centimeter. Density of turns in \(\mathrm{turns/m}\) will be: \( n = 10\ \mathrm{turns/cm} \cdot (\frac{100\ \mathrm{cm}}{1\ \mathrm{m}}) = 1000\ \mathrm{turns/m} \) Now we can calculate B: \( B = (4\pi \times 10^{-7}\ \mathrm{Tm/A})\times (1000\ \mathrm{turns/m})\times(5\ \mathrm{A}) \)

Evaluate the magnetic induction at the axis inside the solenoid

Now, we can evaluate B: \(B = 4\pi \times 10^{-4}\ \mathrm{T}\) This result matches the option (a) \(2\pi \times 10^{-3}\ \mathrm{T}\), so the magnetic induction at the axis inside the solenoid is: \(B = 2\pi \times 10^{-3}\ \mathrm{T}\).